Jones matrices of a mystery device

Question

When considering a Jones matrix

$$J=\ \left( \begin{array}{ccc} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \\ \end{array} \right) $$

I understand that the effect of a device described by this Jones matrix on a linearly polarized light is rotation by angle $\phi$. I identified the corresponding device as a Faraday rotator. I found eigenvalues to be

$$\lambda_1=e^{i\phi} \quad \text{and} \quad \lambda_2=e^{-i\phi}$$

and corresponding eigenvectors as

$$\vec{v}_1=\frac{1}{\sqrt{2}} \left( \begin{array}{ccc} 1 \\ i \\ \end{array} \right) \quad \text{and} \quad \vec{v}_2=\frac{1}{\sqrt{2}}\ \left(\begin{array}{ccc} 1 \\ -i \\ \end{array} \right)\ $$

which have the form of left and right circular polarization states. I found Jones matrix in its own diagonal frame to be

$$J'=\ \left( \begin{array}{ccc} e^{i\phi} & 0 \\ 0 & e^{-i\phi} \\ \end{array} \right)\ $$

When asked to explain rotational effect of a device described by the first Jones matrix by considering an incident polarised wave to be a superposition of eigenpolarizations, how can I approach this?

Answer 1

Since the matrix shown is a rotation matrix, which rotates an input vector by the angle $\phi $, it should be the Jones matrix for a half wave plate.

Click for a list of devices and their corresponding Jones matrix representations.

The half wave plate "flips" the normally incident linear polarization across the fast axis; this means that if the input polarization makes an angle $\theta $ with the fast axis, the total rotation will be $2\theta $.

A detailed analysis is found at Newport's polarization page.

Answer 2

The vectors are $v_1,v_2$ are circularly polarised light, in which the polarisation vector rotates over time. So the matrix $J'$ describes changing the phase relations between $v_1$ and $v_2$. One way to explain the relation would be just to work out the original vector as a superposition of $v_1$ and $v_2$ and then apply $J'$.

Another way to think about it is that $v_1$ and $v_2$ have a polarisation that rotates over time. Linear polarisation in some direction is like the rotation of $v_1$ and $v_2$ cancelling out at a particular angle. And the matrix $J'$ just makes the rotation cancel out at a different angle.

Answer 3

Answering the question first - since you have just worked out the eigenvectors of the matrix representing your mystery device, you can test that the matrix - as by definition - leaves eigenvectors unchanged. A linear combination/superposition of such eigenstates will likewise be unperturbed by such a device.

Regarding the answer by Peter Diehr, no, OP was correct - that matrix represents a Faraday isolator/rotator, with the matrix for a halfwave plate being $$ HWP(\theta) = \begin{pmatrix} \cos(2\theta) & \cos(2\theta) \\ \cos(2\theta) & \sin(2\theta) \end{pmatrix}. $$ (note the lack of minus signs.)